High roots of negative numbers.

**orange gold** · December 28th 2011, 09:15 AM

I was working a simple equation with the sine function, sin(x), and I converted the function to (e^(i*x))-(e^(i*-x))... later on I got to the 10th root of negative 1... so I just converted that to i because (i^10 = -1)... then the equation fell apart and nothing was working after that... does the 10th root of negative 1 have more answers than just "i"? If so, how do I find these roots?

**Plato** · December 28th 2011, 10:01 AM

Originally Posted by orange gold

does the 10th root of negative 1 have more answers than just "i"? If so, how do I find these roots?

There are ten tenth roots of -1.
Let $\sigma=\exp\left(\frac{i\pi}{10}\right)~\&~\xi= \exp \left(\frac{i\pi}{5}\right)$ .

The ten roots are: $\sigma\cdot\xi^k,~k=0,1,\cdots 9$

**orange gold** · December 28th 2011, 12:17 PM

I'm a little unfamiliar with the notation you used. Would the values be complex numbers in the form x+iy? If so how can I find the x and y values? Would cosine and sine be related (assuming this based of your use of pi/n, resembles radian form to me)

**Plato** · December 28th 2011, 12:29 PM

Originally Posted by orange gold

I'm a little unfamiliar with the notation you used. Would the values be complex numbers in the form x+iy? If so how can I find the x and y values? Would cosine and sine be related (assuming this based of your use of pi/n, resembles radian form to me)

The notation is from complex analysis.
It is simple: $\exp \left(i\theta\right)=\cos(\theta)+i\sin(\theta)$ .
Thus $\left[\exp(i\theta)\right]^k=\cos(k\theta)+i\sin(k\theta)$ .
It is easy notation for DeMoivre's Theorem.

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